Physics - EMI/AC Question with Solution | TestHub

P: The quality factor $Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$. Thus (P)-(2). Q: The full width at half maximum (bandwidth) for the power curve is $\Delta \omega =\frac{R}{L}$. Thus (Q)-(3). R: At resonance, the impedance is purely resistive, $Z=R$. The maximum current amplitude is $I_{max}=\frac{V_0}{R}$. Thus (R)-(1). S: The phase difference is given by $\tan \varphi =\frac{\omega L-1/(\omega C)}{R}$. The frequencies ${\omega }_0\pm \frac{R}{2L}$ are the half-power frequencies where $|\omega L-1/(\omega C)|=R$. For $\omega ={\omega }_0+\frac{R}{2L}$, the term $(\omega L-1/(\omega C))$ is positive, so $\tan \varphi =\frac{R}{R}=1$, which means $\varphi =\frac{\pi }{4}$. Thus (S)-(4).